Series $\sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+z} \right)$

Question

If $z$ is an integer, the sum of the series $$\sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+z}\right)$$ is easy since it is a telescoping series. But if $z$ is a fraction, say $z=3/2$, I don't see why the series sums to $$\frac{8}{3}-\ln 4$$ Is there a formula for $z=m/n$, where $m,n$ are positive integers and $n\neq 1$?

Answer 1

For $\vert x \vert < 1$, if we define $f(x)$ as$$f(x)=\sum_{n=0}^{\infty} \left(x^n - x^{n+z}\right)$$ then $$\int_0^1 f(x) dx = \sum_{n=1}^{\infty} \left(\dfrac1n - \dfrac1{n+z}\right)$$ But we have $$f(x) = (1-x^z) \sum_{n=0}^{\infty} x^n = \dfrac{1-x^z}{1-x}$$ Hence, $$\sum_{n=1}^{\infty} \left(\dfrac1n - \dfrac1{n+z}\right) = \int_0^1 \dfrac{1-x^z}{1-x}dx = I(z)$$ For $z=3/2$, we have \begin{align} I(3/2) & = \overbrace{\int_0^1 \dfrac{1-x^{3/2}}{1-x}dx = \int_0^1 \dfrac{1-t^3}{1-t^2} \cdot 2tdt}^{\text{Substitute }x = t^2} = \int_0^1 \dfrac{2t(1+t+t^2)}{1+t}dt\\ & = \int_0^1 \dfrac{2t((1+t)^2-t)}{1+t}dt = \int_0^1 2t(1+t) dt - 2 \int_0^1 \dfrac{t^2}{1+t} dt\\ & = \left(t^2 + \dfrac{2t^3}3\right)_{t=0}^{t=1} - 2 \int_0^1 \dfrac{t^2-1}{t+1}dt - 2 \int_0^1 \dfrac{dt}{1+t}\\ & = \dfrac53 - 2 \left(\dfrac{t^2}2 - t\right)_{t=0}^{t=1}-2\ln(2)\\ & = \dfrac53 - 2 \left(\dfrac12-1\right) - \ln(4) = \dfrac83 - \ln(4) \end{align}

Answer 2

To complete Marvis' (and now Mhenni's) answer let's notice that your series is, up to the Euler $\gamma$ constant ($0.577215\cdots$), the digamma $\psi$ function i.e. $$\psi(z+1)+\gamma=\sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+z}\right)$$

The formula for any fraction is well known too and named Gauss's digamma theorem : $$\psi\left(\frac mk\right)=-\gamma-\ln(2k)-\frac{\pi}2\cot\left(\frac{m\pi}k\right)+2\sum_{n=1}^{\lfloor(k-1)/2\rfloor}\cos\left(\frac{2\pi nm}k\right)\ln\left(\sin\left(\frac{n\pi}k\right)\right)$$ This is valid for $\,0<m<k$. For other fractions (including negative non integer values) use repetitively $\;\displaystyle\psi(x+1)=\frac{1}{x}+\psi(x)$.

Concerning polygamma $\;\psi^{(n)}\left(\frac mk\right)\,$ for $n\in\mathbb{N}\ $ a formula was proposed by K. S. Kölbig in :
"The polygamma function and the derivatives of the cotangent function for rational arguments".

Answer 3

Simple.

$$\begin{align}\sum_{k=1}^\infty\frac1k-\frac1{k+\frac32}&=-2\sum_{k=1}^\infty\frac1{2k+3}-\frac1{2k}\\&=\frac83-2\sum_{k=1}^\infty\frac1{2k-1}-\frac1{2k}\\&=\frac83+2\sum_{k=1}^\infty\frac{(-1)^{k+1}}k\\&=\frac83-2\ln(2)\\&=\frac83-\ln(4)\end{align}$$

This approach works for any $z=\frac12+n,~n\in\Bbb N$.

Answer 4

Recalling the identity of the $\psi$ function

$$ \psi(z)=-\gamma+\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+z}\right)\qquad z\neq0,-1,-2,-3,\ldots, $$

your series can be readily expressed in terms of the $\psi$ function

$$ \sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+z}\right)= \psi(z+1)+\gamma. $$

Note that, for $z=\frac{m}{n},\, m,n\in \mathbb{N}$, you will not have a problem, since the singularities of the $\psi$ function are at $z=0,-1,-2,\dots.$